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Lostsunrise [7]

3 years ago

10

Mario brought a 1 gallon can of paint to paint his room. He used 0.5 gal for the fist coat and 0.4 gal for the second coat. How

much paint did he have left

2 answers:

emmainna [20.7K]3 years ago

7 0

It would be 1/10 OR 0.1

Lady_Fox [76]3 years ago

5 0

0.1 left because 1 gallon, .4+.5=.9. 1-.9= .1

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Answer: A.

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julia-pushkina [17]

Answer:

Approximately $101\; \rm ft$ (assuming that the height of the base of the hill is the same as that of the observer.)

Step-by-step explanation:

Refer to the diagram attached.

Let $\rm O$ denote the observer.
Let $\rm A$ denote the top of the tree.
Let $\rm R$ denote the base of the tree.
Let $\rm B$ denote the point where line $\rm AR$ (a vertical line) and the horizontal line going through $\rm O$ meets. $\angle \rm B\hat{A}R = 90^\circ$ .

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Angle of elevation of the base of the tree as it appears to the observer: $\angle \rm B\hat{O}R = 51^\circ$ .
Angle of elevation of the top of the tree as it appears to the observer: $\angle \rm B\hat{O}A = 57^\circ$ .

Let the length of segment $\rm BR$ (vertical distance between the base of the tree and the base of the hill) be $x\; \rm ft$ .

The question is asking for the length of segment $\rm AB$ . Notice that the length of this segment is $\mathrm{AB} = (x + 20)\; \rm ft$ .

The length of segment $\rm OB$ could be represented in two ways:

In right triangle $\rm \triangle OBR$ as the side adjacent to $\angle \rm B\hat{O}R = 51^\circ$ .
In right triangle $\rm \triangle OBA$ as the side adjacent to $\angle \rm B\hat{O}A = 57^\circ$ .

For example, in right triangle $\rm \triangle OBR$ , the length of the side opposite to $\angle \rm B\hat{O}R = 51^\circ$ is segment $\rm BR$ . The length of that segment is $x\; \rm ft$ .

$\begin{aligned}\tan{\left(\angle\mathrm{B\hat{O}R}\right)} = \frac{\,\rm {BR}\,}{\,\rm {OB}\,} \; \genfrac{}{}{0em}{}{\leftarrow \text{opposite}}{\leftarrow \text{adjacent}}\end{aligned}$ .

Rearrange to find an expression for the length of $\rm OB$ (in $\rm ft$ ) in terms of $x$ :

$\begin{aligned}\mathrm{OB} &= \frac{\mathrm{BR}}{\tan{\left(\angle\mathrm{B\hat{O}R}\right)}} \\ &= \frac{x}{\tan\left(51^\circ\right)}\approx 0.810\, x\end{aligned}$ .

Similarly, in right triangle $\rm \triangle OBA$ :

$\begin{aligned}\mathrm{OB} &= \frac{\mathrm{AB}}{\tan{\left(\angle\mathrm{B\hat{O}A}\right)}} \\ &= \frac{x + 20}{\tan\left(57^\circ\right)}\approx 0.649\, (x + 20)\end{aligned}$ .

Equate the right-hand side of these two equations:

$0.810\, x \approx 0.649\, (x + 20)$ .

Solve for $x$ :

$x \approx 81\; \rm ft$ .

Hence, the height of the top of this tree relative to the base of the hill would be $(x + 20)\; {\rm ft}\approx 101\; \rm ft$ .

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